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            在
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             5 年, 6 月 之前
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               2010年12月2日 下午3:22
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               <strong>
                题
                <bblatex>
                 \mathbb{R}
                </bblatex>
                中存在一个子集，其下确界(infimum)大于上确界(supremum)。
               </strong>
               <br/>
               解: 站在单点的角度，注意到任何单点都同时是一个集合的inf和其补集的sup。设想用sup和inf分别描述”无穷大比任何实数都大”这句话分别应为:
               <bblatex>
                \infty=\sup\mathbb{R}
               </bblatex>
               和
               <bblatex>
                \infty=\inf(\mathbb{R}-\mathbb{R})=\inf\emptyset
               </bblatex>
               <br/>
               同理
               <bblatex>
                \sup\emptyset=-\infty
               </bblatex>
               <br/>
               于是，在
               <bblatex>
                \mathbb{R}
               </bblatex>
               中，对空集来说，其上确界小于下确界。
               <br/>
               从另一个角度看，由于任何数都是空集的上界，所以最小上界是
               <bblatex>
                -\infty
               </bblatex>
               。这是因为按照定义，如果对每一个
               <bblatex>
                a\in E
               </bblatex>
               都有
               <bblatex>
                a&lt;m
               </bblatex>
               ，则
               <bblatex>
                m
               </bblatex>
               为
               <bblatex>
                E
               </bblatex>
               的一个上界，而按照逻辑，前提不成立的命题为真命题。
               <br/>
               由于任何数也是空集的下界，所以最大下界为
               <bblatex>
                \infty
               </bblatex>
               .
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               2010年12月2日 下午6:39
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               <strong>
                题 举一个按分布收敛但不按概率收敛的例子。
               </strong>
               <br/>
               解 设
               <bblatex>
                X \sim N\left( {0,1} \right)
               </bblatex>
               且
               <bblatex>
                \forall n \in \mathbb{N}, X_n=-X
               </bblatex>
               ，则
               <br/>
               <bblatex>
                \mathbb{P}\left( {{X_n} \leqslant x} \right) = \mathbb{P}\left( { – X \leqslant x} \right) = \mathbb{P}\left( {X \geqslant  – x} \right) = \mathbb{P}\left( {X \leqslant x} \right)
               </bblatex>
               ，故
               <bblatex>
                {X_n}\xrightarrow{\mathcal{D}}X
               </bblatex>
               .
               <br/>
               但显然
               <bblatex>
                X_n
               </bblatex>
               不按概率收敛，用符号写下来就是
               <br/>
               <bblatex>
                \forall \varepsilon  &gt; 0,\mathbb{P}\left( {\left| {{X_n} – X} \right| &lt; \varepsilon } \right) = \mathbb{P}\left( {\left| X \right| &lt; \frac{\varepsilon }{2}} \right) = 2\Phi \left( {\frac{\varepsilon }{2}} \right) – 1
               </bblatex>
               不随n动，从而不会趋于1。
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               2010年12月2日 下午7:01
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              <p>
               <strong>
                题 举一个按概率收敛但不是几乎确定地收敛的例子。
               </strong>
               <br/>
               解 作
               <bblatex>
                {X_n} = X + {W_n}
               </bblatex>
               ,其中
               <bblatex>
                W_n
               </bblatex>
               完全独立，且
               <bblatex>
                \begin{array}{*{20}{c}}  {\mathbb{P}\left( {{W_n} = {2^{ – n}}} \right) = 1 – \frac{1}{n}}&amp;{\mathbb{P}\left( {{W_n} = 1000n} \right) = \frac{1}{n}}\end{array}
               </bblatex>
               <br/>
               <bblatex>
                \mathbb{P}\left( {\left| {{X_n} – X} \right| &lt; \varepsilon } \right)= \mathbb{P}\left( {\left| {{W_n}} \right| &lt; \varepsilon } \right) = \mathbb{P}\left( {{W_n} &lt; \varepsilon } \right)\xrightarrow{{n \to \infty }}1 – \frac{1}{n} \to 1
               </bblatex>
               <br/>
               <bblatex>
                \therefore {X_n}\xrightarrow{\mathbb{P}}X
               </bblatex>
               <br/>
               另一方面，
               <br/>
               <bblatex>
                \begin{gathered} \mathbb{P}\left( {\left| {{X_n} – X} \right| &lt; \varepsilon ,\forall n \geqslant m} \right) = \mathbb{P}\left( {{W_m} &lt; \varepsilon ,{W_{m + 1}} &lt; \varepsilon , \cdots } \right) \hfill \\ = \left( {1 – \frac{1}{m}} \right)\left( {1 – \frac{1}{{m + 1}}} \right) \cdots  \leqslant \left( {1 – \frac{1}{m}} \right)\left( {1 – \frac{1}{{m + 1}}} \right) \cdots \left( {1 – \frac{1}{M}} \right)\left( {\forall M &gt; m} \right) \hfill \\ = \frac{{m – 1}}{m}\frac{m}{{m + 1}}\frac{{m + 1}}{{m + 2}} \cdots \frac{{M – 1}}{M} = \frac{{m – 1}}{M}\xrightarrow{{M \to \infty }}0 \hfill \\ \end{gathered}
               </bblatex>
               <bblatex>
                = \frac{{m – 1}}{m}\frac{m}{{m + 1}}\frac{{m + 1}}{{m + 2}} \cdots \frac{{M – 1}}{M} = \frac{{m – 1}}{M}\xrightarrow{{M \to \infty }}0
               </bblatex>
               <br/>
               故
               <bblatex>
                X_n
               </bblatex>
               不几乎确定收敛到
               <bblatex>
                X
               </bblatex>
               .
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               2010年12月3日 上午11:48
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              <p>
               第一题是有问题的吧。虽然都叫空集，但两个空集可能指的完全不同的两个东西。集合
               <bblatex>
                (a,a+\frac{1}{n})
               </bblatex>
               的上确界是
               <bblatex>
                a+\frac{1}{n}
               </bblatex>
               ，让n趋于无穷，则可以说空集上确界是a，但是只能得出结论说这个特殊的空集它的上确界是a，而不能得出结论说所有的空集上确界是a。
              </p>
              <p>
               这个证明的确证明了存在某个空集它的下确界是
               <bblatex>
                \infty
               </bblatex>
               ，但是在同理的时候，却是证明了存在另外一个完全不相关的空集其上确界是
               <bblatex>
                -\infty
               </bblatex>
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               2010年12月3日 下午12:10
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              <p>
               回复 第4楼 的 lanfeng：这个要讨论比较深了。首先从逻辑角度上讲，”不含任何元素的集合”本身蕴含着空集的唯一性。其次，“lim和sup两个操作符何时可以交换”是一个没有“空集的唯一性”来得基本的问题。事实上，通常定义
               <bblatex>
                \mathbb{R}
               </bblatex>
               中空集的上确界为
               <bblatex>
                -\infty
               </bblatex>
               ，下确界为
               <bblatex>
                \infty
               </bblatex>
               .
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               2010年12月3日 下午12:17
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               6 楼
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              <p>
               回复 第5楼 的 easttiger：空集无所谓上界和下界了，因此可以随意指定，因此才有1楼的“怪论”。这个也是平凡的问题，随便按常规想法定义就OK了，不会影响什么别的。
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               2010年12月3日 下午1:21
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              <p>
               <strong>
                题 对只含一个一维参数
                <bblatex>
                 \theta_0
                </bblatex>
                的一元分布
                <bblatex>
                 f(x|\theta_0)
                </bblatex>
                的简单的情况证明极大似然估计量
                <bblatex>
                 {{\hat \theta }_n}
                </bblatex>
                为
                <bblatex>
                 \theta_0
                </bblatex>
                的一致(consistent)估计量。即证明
                <bblatex>
                 {{\hat \theta }_n}\xrightarrow{\mathbb{P}}{\theta _0}
                </bblatex>
                .
               </strong>
               <br/>
               证
               <br/>
               I. 基本设置。
               <br/>
               样本:
               <bblatex>
                {X_i}\mathop  \sim \limits^{iid} f\left( { \cdot \left| {{\theta _0}} \right.} \right)
               </bblatex>
               <br/>
               Score function:
               <bblatex>
                {S_n}\left( \theta  \right): = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{\partial }{{\partial \theta }}\ln f\left( {{X_i}\left| \theta  \right.} \right)}  = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{f’\left( {{X_i}\left| \theta  \right.} \right)}}{{f\left( {{X_i}\left| \theta  \right.} \right)}}}
               </bblatex>
               <br/>
               MLE:
               <bblatex>
                {{\hat \theta }_n}
               </bblatex>
               为
               <bblatex>
                {S_n}\left( \theta  \right)
               </bblatex>
               的根。
               <br/>
               II.
               <a href="http://en.wikipedia.org/wiki/Maximum_likelihood#Consistency" rel="nofollow">
                证明开始
               </a>
               <br/>
               令
               <bblatex>
                M\left( \theta  \right): = \mathbb{E}\left[ {\ln f\left( {{X_1}\left| \theta  \right.} \right)} \right]
               </bblatex>
               ，则以下两则引理成立:
               <br/>
               (1)
               <bblatex>
                \arg \max M\left( \theta  \right) = {\theta _0}
               </bblatex>
               <br/>
               (2)
               <bblatex>
                {S_n}\left( \theta  \right)\xrightarrow{\mathbb{P}}M’\left( \theta  \right)
               </bblatex>
               <br/>
               证明(1):
               <br/>
               <bblatex>
                M\left( \theta  \right) – M\left( {{\theta _0}} \right) = \mathbb{E}\left[ {\ln \frac{{f\left( {{X_1}\left| \theta  \right.} \right)}}{{f\left( {{X_1}\left| {{\theta _0}} \right.} \right)}}} \right] \leqslant \ln \mathbb{E}\left[ {\frac{{f\left( {{X_1}\left| \theta  \right.} \right)}}{{f\left( {{X_1}\left| {{\theta _0}} \right.} \right)}}} \right]
               </bblatex>
               （琴生不等式）
               <br/>
               <bblatex>
                \mathbb{E}\left[ {\frac{{f\left( {{X_1}\left| \theta  \right.} \right)}}{{f\left( {{X_1}\left| {{\theta _0}} \right.} \right)}}} \right] = \int {\frac{{f\left( {{x_1}\left| \theta  \right.} \right)}}{{f\left( {{x_1}\left| {{\theta _0}} \right.} \right)}}f\left( {{x_1}\left| {{\theta _0}} \right.} \right)d} {x_1} = 1
               </bblatex>
               <br/>
               故
               <bblatex>
                \therefore M\left( \theta  \right) \leqslant M\left( {{\theta _0}} \right)
               </bblatex>
               .
               <br/>
               证明(2):
               <br/>
               <bblatex>
                M’\left( \theta  \right) = \frac{\partial }{{\partial \theta }}\mathbb{E}\left[ {\ln f\left( {{X_1}\left| \theta  \right.} \right)} \right] = \mathbb{E}\left[ {\frac{\partial }{{\partial \theta }}\ln f\left( {{X_1}\left| \theta  \right.} \right)} \right] = \mathbb{E}\left( {{S_n}\left( \theta  \right)} \right)
               </bblatex>
               ,
               <br/>
               于是，由弱大数定律，
               <bblatex>
                {S_n}\left( \theta  \right)\xrightarrow{\mathbb{P}}M’\left( \theta  \right)
               </bblatex>
               .
               <br/>
               目标:
               <bblatex>
                \left( {\forall \varepsilon  &gt; 0} \right)\left[ {\mathbb{P}\left( {\left| {{{\hat \theta }_n} – {\theta _0}} \right| &lt; \varepsilon } \right) \to 1} \right]
               </bblatex>
               <br/>
               目标的证明:
               <br/>
               <bblatex>
                \forall \varepsilon  &gt; 0
               </bblatex>
               ,
               <br/>
               <bblatex>
                \begin{gathered} {S_n}\left( {{\theta _0} – \varepsilon } \right)\xrightarrow{\mathbb{P}}M’\left( {{\theta _0} – \varepsilon } \right) &gt; 0 \hfill \\ {S_n}\left( {{\theta _0} + \varepsilon } \right)\xrightarrow{\mathbb{P}}M’\left( {{\theta _0} + \varepsilon } \right) &lt; 0 \hfill \\ \end{gathered}
               </bblatex>
               <br/>
               <bblatex>
                \begin{gathered}\mathbb{P}\left\{ {\left| {{S_n}\left( {{\theta _0} – \varepsilon } \right) – M’\left( {{\theta _0} – \varepsilon } \right)} \right| \geqslant \frac{1}{2}M’\left( {{\theta _0} – \varepsilon } \right){\text{ or }}\left| {{S_n}\left( {{\theta _0} + \varepsilon } \right) – M’\left( {{\theta _0} + \varepsilon } \right)} \right| \geqslant  – \frac{1}{2}M’\left( {{\theta _0} + \varepsilon } \right)} \right\} \to 0 \hfill \\  \mathbb{P}\left\{ {\left| {{S_n}\left( {{\theta _0} – \varepsilon } \right) – M’\left( {{\theta _0} – \varepsilon } \right)} \right| &lt; \frac{1}{2}M’\left( {{\theta _0} – \varepsilon } \right){\text{ \&amp;  }}\left| {{S_n}\left( {{\theta _0} + \varepsilon } \right) – M’\left( {{\theta _0} + \varepsilon } \right)} \right| &lt;  – \frac{1}{2}M’\left( {{\theta _0} + \varepsilon } \right)} \right\} \to 1 \hfill \\ \end{gathered}
               </bblatex>
               <br/>
               <bblatex>
                \begin{gathered} \mathbb{P}\left\{ {{S_n}\left( {{\theta _0} – \varepsilon } \right) &gt; \frac{1}{2}M’\left( {{\theta _0} – \varepsilon } \right){\text{ \&amp;  }}{S_n}\left( {{\theta _0} + \varepsilon } \right) &lt; \frac{1}{2}M’\left( {{\theta _0} + \varepsilon } \right)} \right\} \to 1 \hfill \\  \mathbb{P}\left\{ {{S_n}\left( {{\theta _0} – \varepsilon } \right) &gt; 0{\text{ \&amp;   }}{S_n}\left( {{\theta _0} + \varepsilon } \right) &lt; 0} \right\} \to 1 \hfill \\ \end{gathered}
               </bblatex>
               <br/>
               <bblatex>
                \mathbb{P}\left\{ {{\theta _0} – \varepsilon  \leqslant {{\hat \theta }_n} \leqslant {\theta _0} + \varepsilon } \right\} \to 1
               </bblatex>
               <br/>
               故
               <bblatex>
                {{\hat \theta }_n}\xrightarrow{\mathbb{P}}{\theta _0}
               </bblatex>
              </p>
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               2010年12月3日 下午1:56
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               8 楼
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              <p>
               回复 第6楼 的 cloud_wei：可以换个角度看，由于任何数都是空集的上界，所以最小上界是
               <bblatex>
                -\infty
               </bblatex>
               。事实上按照上确界的原始定义”最小的上界”来看，上面那个
               <bblatex>
                (a,a+1/n)
               </bblatex>
               的例子最多(假设limsup=suplim前提下)说明
               <bblatex>
                a
               </bblatex>
               是
               <bblatex>
                \emptyset
               </bblatex>
               的一个上界。
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            <div class="bbp-reply-header" id="post-311596">
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               2010年12月3日 下午5:19
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              <p>
               <strong>
                接上题, 继续证明MLE
                <bblatex>
                 \hat{\theta_n}
                </bblatex>
                的极限分布为
                <bblatex>
                 N(\theta_0, \frac{1}{nI_0})
                </bblatex>
                , 其中
                <bblatex>
                 {I_0} = \mathbb{V} \left[ {\frac{\partial }{{\partial \theta }}\ln f\left( {{X_1}\left| \theta  \right.} \right)\left| {_{\theta  = {\theta _0}}} \right.} \right]=  – \mathbb{E}\left[ {\frac{{{\partial ^2}}}{{\partial {\theta ^2}}}\ln f\left( {{X_1}\left| \theta  \right.} \right)\left| {_{\theta  = {\theta _0}}} \right.} \right]
                </bblatex>
                为Fisher Information Value.
               </strong>
               <br/>
               证
               <br/>
               引理:
               <bblatex>
                {S’_n}\left( {{\theta _0}} \right)\xrightarrow{\mathbb{P}} – {I_0}
               </bblatex>
               <br/>
               证明引理:
               <br/>
               <bblatex>
                \begin{gathered} {{S’}_n}\left( {{\theta _0}} \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\partial ^2}}}{{\partial {\theta ^2}}}\ln f\left( {{X_i}\left| \theta  \right.} \right)\left| {_{\theta  = {\theta _0}}} \right.}  \hfill \\ \end{gathered}
               </bblatex>
               <br/>
               于是，由弱大数定律，
               <bblatex>
                {{S’}_n}\left( {{\theta _0}} \right)\xrightarrow{\mathbb{P}} – {I_0}
               </bblatex>
               <br/>
               <a href="http://en.wikipedia.org/wiki/Maximum_likelihood#Asymptotic_normality" rel="nofollow">
                证明原命题
               </a>
               :
               <br/>
               <bblatex>
                \begin{gathered}  0 = {S_n}\left( {{{\hat \theta }_n}} \right) = {S_n}\left( {{\theta _0}} \right) + \left( {{{\hat \theta }_n} – {\theta _0}} \right){{S’}_n}\left( {{\theta _0}} \right) + \frac{1}{2}{\left( {{{\hat \theta }_n} – {\theta _0}} \right)^2}{O_p}\left( 1 \right) \hfill \\  {{\hat \theta }_n} – {\theta _0} = \frac{{{S_n}\left( {{\theta _0}} \right)}}{{ – {{S’}_n}\left( {{\theta _0}} \right) – \left( {{{\hat \theta }_n} – {\theta _0}} \right){O_p}\left( 1 \right)}} \hfill \\  {S_n}\left( {{\theta _0}} \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{\partial }{{\partial \theta }}f\left( {{X_i}\left| \theta  \right.} \right)\left| {_{\theta  = {\theta _0}}} \right.}  \hfill \\  \mathbb{E}\left[ {{S_n}\left( {{\theta _0}} \right)} \right]\xrightarrow{\mathbb{P}}\mathbb{E}\left[ {{S_n}\left( {{{\hat \theta }_n}} \right)} \right] = 0 \hfill \\  \mathbb{V}\left[ {{S_n}\left( {{\theta _0}} \right)} \right] = \frac{1}{n}\mathbb{V}\left[ {\frac{\partial }{{\partial \theta }}f\left( {{X_1}\left| \theta  \right.} \right)\left| {_{\theta  = {\theta _0}}} \right.} \right] = \frac{{{I_0}}}{n} \hfill \\ \end{gathered}
               </bblatex>
               <br/>
               <bblatex>
                \therefore {S_n}\left( {{\theta _0}} \right)\xrightarrow{\mathcal{D}}N\left( {0,\frac{{{I_0}}}{n}} \right)
               </bblatex>
               (由
               <a href="http://en.wikipedia.org/wiki/Central_limit_theorem" rel="nofollow">
                CLT
               </a>
               )
               <br/>
               <bblatex>
                \therefore {{\hat \theta }_n} – {\theta _0}\xrightarrow{\mathcal{D}}N\left( {0,\frac{1}{{n{I_0}}}} \right)
               </bblatex>
               (由
               <a href="http://en.wikipedia.org/wiki/Slutsky’s_theorem" rel="nofollow">
                Slutsky
               </a>
               )
              </p>
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               2010年12月3日 下午6:23
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              <p>
               <strong>
                题 证明:基于iid样本
                <bblatex>
                 \left\{ {{X_i}:i = 1, \cdots ,n} \right\}
                </bblatex>
                的经验分布函数
                <bblatex>
                 {\hat F_n}\left( t \right)
                </bblatex>
                是总体分布函数
                <bblatex>
                 F(t)
                </bblatex>
                在取定点
                <bblatex>
                 t
                </bblatex>
                的一致且无偏(consistent and unbiased)估计量。接着用CLT和Slutsky证明
                <a href="http://en.wikipedia.org/wiki/Empirical_distribution_function#Asymptotic_properties" rel="nofollow">
                 其极限分布
                </a>
                之一为
                <bblatex>
                 {{\hat F}_n}\left( t \right)\xrightarrow{\mathcal{D}}N\left( {F\left( t \right),\frac{1}{n}{{\hat F}_n}\left( t \right)\left( {1 – {{\hat F}_n}\left( t \right)} \right)} \right)
                </bblatex>
                .
               </strong>
               <br/>
               证
               <br/>
               I.基本设置
               <br/>
               样本:
               <bblatex>
                X_1,X_2,\cdots,X_n \mathop  \sim \limits^{iid}  F(t)
               </bblatex>
               <br/>
               经验分布函数的定义：
               <bblatex>
                {\hat F_n}\left( t \right): = \frac{1}{n} \sum\limits_{i = 1}^n {1\left( {{X_i} \leqslant t} \right)}
               </bblatex>
               <br/>
               II.证明开始
               <br/>
               (1)无偏性
               <br/>
               <bblatex>
                \forall t \in \mathbb{R},\mathbb{E}\left[ {{{\hat F}_n}\left( t \right)} \right] = \mathbb{E}\left[ {\frac{1}{n}\sum\limits_{i = 1}^n {1\left( {{X_i} \leqslant t} \right)} } \right] = \frac{1}{n}\sum\limits_{i = 1}^n {\mathbb{P}\left( {{X_i} \leqslant t} \right)}  = F\left( t \right)
               </bblatex>
               <br/>
               (2)一致性
               <br/>
               由
               <a href="http://en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law" rel="nofollow">
                WLLN
               </a>
               ,
               <bblatex>
                \forall t \in \mathbb{R},{{\hat F}_n}\left( t \right)\xrightarrow{\mathbb{P}}F\left( t \right)
               </bblatex>
               <br/>
               (3)完全由样本确定参数的极限分布。
               <br/>
               首先推导基于CLT的极限分布
               <br/>
               <bblatex>
                \forall t \in \mathbb{R},\mathbb{V}\left[ {{{\hat F}_n}\left( t \right)} \right] = \frac{1}{n}\mathbb{V}\left[ {1\left( {{X_1} \leqslant t} \right)} \right] = \frac{1}{n}F\left( t \right)\left( {1 – F\left( t \right)} \right)\xrightarrow{\mathcal{D}}0
               </bblatex>
               <br/>
               <bblatex>
                \forall t \in \mathbb{R},{{\hat F}_n}\left( t \right)\xrightarrow{\mathcal{D}}N\left( {F\left( t \right),\frac{1}{n}F\left( t \right)\left( {1 – F\left( t \right)} \right)} \right)
               </bblatex>
               (由CLT)
               <br/>
               原命题等价于
               <br/>
               <bblatex>
                \frac{{{{\hat F}_n}\left( t \right) – F\left( t \right)}}{{\sqrt {\frac{1}{n}{{\hat F}_n}\left( t \right)\left( {1 – {{\hat F}_n}\left( t \right)} \right)} }}\xrightarrow{\mathcal{D}}N\left( {0,1} \right)
               </bblatex>
              </p>
              <p>
               由(2)的一致性结论，
               <bblatex>
                \frac{1}{n}{{\hat F}_n}\left( t \right)\left( {1 – {{\hat F}_n}\left( t \right)} \right)\xrightarrow{\mathbb{P}}\frac{1}{n}F\left( t \right)\left( {1 – F\left( t \right)} \right)
               </bblatex>
               ;
               <br/>
               <bblatex>
                \therefore \frac{{{{\hat F}_n}\left( t \right) – F\left( t \right)}}{{\sqrt {\frac{1}{n}{{\hat F}_n}\left( t \right)\left( {1 – {{\hat F}_n}\left( t \right)} \right)} }}\xrightarrow{\mathcal{D}}N\left( {0,1} \right)
               </bblatex>
               (Slutsky)
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